The values of $$\lambda$$ and $$\mu$$ such that the system of equations $$x + y + z = 6$$, $$3x + 5y + 5z = 26$$ and $$x + 2y + \lambda z = \mu$$ has no solution, are:

The system is: $$x + y + z = 6$$ $$-(1)$$, $$3x + 5y + 5z = 26$$ $$-(2)$$, $$x + 2y + \lambda z = \mu$$ $$-(3)$$.

Form the augmented matrix and row-reduce:

$$\begin{pmatrix} 1 & 1 & 1 & 6 \\ 3 & 5 & 5 & 26 \\ 1 & 2 & \lambda & \mu \end{pmatrix}$$

Apply $$R_2 \to R_2 - 3R_1$$ and $$R_3 \to R_3 - R_1$$:

$$\begin{pmatrix} 1 & 1 & 1 & 6 \\ 0 & 2 & 2 & 8 \\ 0 & 1 & \lambda - 1 & \mu - 6 \end{pmatrix}$$

Apply $$R_3 \to R_3 - \frac{1}{2}R_2$$:

$$\begin{pmatrix} 1 & 1 & 1 & 6 \\ 0 & 2 & 2 & 8 \\ 0 & 0 & \lambda - 2 & \mu - 10 \end{pmatrix}$$

For no solution, the last row must give an inconsistency: $$0 \cdot z = (\mu - 10) \neq 0$$. This requires $$\lambda - 2 = 0$$ (so the coefficient of $$z$$ vanishes) and $$\mu - 10 \neq 0$$ (so the right side is nonzero).

Therefore $$\lambda = 2$$ and $$\mu \neq 10$$.

The answer is $$\lambda = 2, \mu \neq 10$$, which is Option D.